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Find the values of a, b, c and k.

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anemone

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Feb 14, 2012
3,755
Hi,

The problem stated below has me stumped. I've been trying to solve it for some times but I kept ended up with zero equals zero kind of silly result and I'd appreciate it if someone could show me some ideas on how to crack it.

Thanks in advance.

Problem:

Let \(\displaystyle a,\;b,\;c\) be roots of equation \(\displaystyle x^3-6x^2+kx+k=0\) and \(\displaystyle (a-1)^3+(b-2)^3+(c-3)^3=0\). Find a, b, c and k.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi,

The problem stated below has me stumped. I've been trying to solve it for some times but I kept ended up with zero equals zero kind of silly result and I'd appreciate it if someone could show me some ideas on how to crack it.

Thanks in advance.

Problem:

Let \(\displaystyle a,\;b,\;c\) be roots of equation \(\displaystyle x^3-6x^2+kx+k=0\) and \(\displaystyle (a-1)^3+(b-2)^3+(c-3)^3=0\). Find a, b, c and k.
Since $a,b,c$ are roots of $x^3-6x^2+kx+k=0$, we have $a+b+c=6\Rightarrow (a-1)+(b-2)+(c-3)=0$. This means $(a-1)^3+(b-2)^3+(c-3)^3=3(a-1)(b-2)(c-3)$. So you get $(a-1)(b-2)(c-3)=0$.
Now
We already have $ab+bc+ca=k$ and $abc=-k$. This should help.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Since $a,b,c$ are roots of $x^3-6x^2+kx+k=0$, we have $a+b+c=6\Rightarrow (a-1)+(b-2)+(c-3)=0$. This means $(a-1)^3+(b-2)^3+(c-3)^3=3(a-1)(b-2)(c-3)$.
Thanks caffeinemachine for the quick reply..

I'm astonished to see another relation exists between $a,b,c$, i.e $(a-1)+(b-2)+(c-3)=0$. I must thank you for that!:)

If I may start from $(a-1)+(b-2)+(c-3)=0$, I get:

$(a-1)+(b-2)=-(c-3)$

$((a-1)+(b-2))^3=(-(c-3))^3$

$(a-1)^3+3(a-1)^2(b-2)+3(a-1)(b-2)^2+(b-2)^3=-(c-3)^3$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)((a-1)+(b-2))=0$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)(a+b-3)=0$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)(6-c-3)=0$

$(a-1)^3++(b-2)^3+(c-3)^3+3(a-1)(b-2)(3-c)=0$

Since we're told that $(a-1)^3++(b-2)^3+(c-3)^3=0$, it means $3(a-1)(b-2)(3-c)=0$, or

$a=1,\;b=2,\;c=3$.

Since $ab+bc+ac=k$ and $abc=-k$, we noticed that $a=1,\;b=2,\;c=3$ can't be true at the same time. In other words, we need to investigate for each of the cases.

Case 1 (When \(\displaystyle a=1\)):

From the Newton Identities we obtained the following equations:

\(\displaystyle a+b+c=6\)

\(\displaystyle a^2+b^2+c^2=36-2k\)

\(\displaystyle a^3+b^3+c^3=216-21k\)

Substitute \(\displaystyle a=1\) into each of the equations above leads us to

\(\displaystyle b+c=5\)

\(\displaystyle b^2+c^2=35-2k\)

\(\displaystyle b^3+c^3=215-21k\)

And by using the identity \(\displaystyle b^3+c^3=(b+c)(b^2-bc+b^2)\) and \(\displaystyle (1)(bc)=-k\), we get:

\(\displaystyle k=\frac{5}{2},\;b=\frac{5-\sqrt{35}}{2},\;c=\frac{5+\sqrt{35}}{2}\)

And we repeat all these for the second and third cases to find

\(\displaystyle a=\frac{2(3+\sqrt{5})}{2},\;b=2,\;c=\frac{2(3-\sqrt{5})}{2},\;k=\frac{16}{3}\)

and

\(\displaystyle a=\frac{3(1+\sqrt{2})}{2}\;c=\frac{3(1-\sqrt{2})}{2},\;c=3,\;k=\frac{27}{4}\)

Yeah! It seems to me now that this is an easy problem!:eek:

Thank you enormously, caffeinemachine!:)
 
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caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
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kaliprasad

Well-known member
Mar 31, 2013
1,322
I proceed slight differently

We have sum of the roots
a+b+c = 6

so (a + b+ c- 6) = 0
or (a-1)+(b-2)+(c-3) = 0

so (a-1)^3 + (b-2)^3 + (c-3)^3= 3(a-1)(b-2)(c-3) ( as x+ y + z = 0 => x^3+y^3 + z^3 = 3xyz

so 3(a-1)(b-2)(c-3) = 0 => a= 1 or b= 2 or c = 3

f(x) = x^3−6x^2+kx+k

taking 1 as a root we get

f(x)= 1-6+ k+ k = 0 or k = 5/2

so we get x^3-6x^2 + 5/2 x + 5/2 = 0

or 2x^3 – 12x^2 + 5 x + 5 = 0

factoring we get (x-1)(2x^2 – 10 x - 5) = 0

we can solve (2x^2 – 10 x - 5)= 0 to get (10+/-√(140))/4 so b = (5+ √35)/2 and c = (5 - √35)/2

another solution b = 5- √35)/2 and c = (5 - √35)/2

similarly other 4 sets of 3 roots can be found
b = 2 shall give 2 sets and c = 3 shall give 2 sets

( kindly note that as we have been given the condition

(b = x1, c= x2) is different from ( b = x2, c= x1))
 
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